If you try to hammer a book into your wall, will it go in? Of course not. But a nail upon hammering readily penetrates the wall. This is because there is more concentration of force at the tip of the nail when hammered than on a book’s surface. Pressure is the force applied per unit area perpendicular to the surface of an object. Pressure is expressed using various units, most of which are derived from a unit of force divided by a unit of area; the SI unit of pressure, the pascal (Pa), for example, is one newton per square metre (N/m2). Another unit of pressure is pounds per square inch. Read on to know more fascinating facts and examples of pressure.

Learn All the Concepts of Pressure

What is Pressure?

To understand pressure, we need to understand the concept of thrust. Thrust is the amount of force acting perpendicular to a surface.

This will help us understand why the sharp edge of a knife is used to cut fruits and vegetables, why the pointed end of the nail is driven to the wall, and the blunt end is hammered, etc.

Definition and Formula of Pressure

Pressure is the force or thrust exerted per unit area. Thus, it is the thrust acting per unit surface area.

From the formula, it is clear that with the increase and decrease in thrust and are the pressure will increase and decrease respectively.

If we keep a brick on a piece of sponge, the sponge will get compressed where the brick is making the contact with it. If the area in contact is more, then the sponge will be compressed less and if the area in contact is less, then the sponge will be compressed more. Thus, pressure can be increased or decreased by changing the area of contact with the object.

Units of Pressure

The SI unit of force is newton \(\left( {\rm{N}} \right)\) and the SI unit of area is metre square \(\left( {{{\rm{m}}^{\rm{2}}}} \right)\).
So, the SI unit of pressure is \({\rm{N}}\,{{\rm{m}}^{ – 2}}.\) The SI unit of pressure is called pascal \(\left( {{\rm{Pa}}} \right)\) in honour of Blaise Pascal. Hence, \({\rm{1}}{\mkern 1mu} {\rm{Pa = 1}}{\mkern 1mu} {\rm{N}}{\mkern 1mu} {{\rm{m}}^{ – 2}}\)
One pascal is defined as the pressure exerted on a surface of area \({\rm{1}}{\mkern 1mu} \,{{\rm{m}}^{\rm{2}}}\)  by a thrust of \({\rm{1}}{\mkern 1mu} \,{\rm{N}}\)  acting on it.

Other units of pressure are

1. \({\rm{k}}{{\rm{g}}_f}{{\rm{m}}^{ – 2}}\) where \({\rm{k}}{{\rm{g}}_f}{{\rm{m}}^{{\rm{ – 2}}}}{\rm{ = }}\,{\rm{10}}\,{\rm{Pa}}\) and
2. bar where \(1\,{\rm{bar}} = {10^5}\,{\rm{Pa}}.\)

Solved Examples

Q.1: A force of \({\rm{300}}\,{\rm{N}}\) is uniformly distributed over an area of \({\rm{60}}\,{{\rm{m}}^{\rm{2}}}{\rm{.}}\) Find the pressure in pascals.
Sol: Given that,
The force is \(F\,{\rm{ = }}\,{\rm{300}}\,{\rm{N}}\)
The area is \(A\,{\rm{ = }}\,6{\rm{0}}\,{{\rm{m}}^2}\)
The pressure is \(P = \frac{F}{A} = \frac{{300}}{{60}} = 5\;{\rm{Pa}}\)
Thus, the pressure exerted on the area in pascals is \({\rm{5}}\,{\rm{Pa}}{\rm{.}}\)

Q.2. How much force should be applied on an area of \({\rm{2}}\,{{\rm{m}}^{\rm{2}}}\) to get a pressure of \({\rm{20}}\,{\rm{Pa?}}\)
Sol: Given that,
The given area is \(A{\rm{ = 2}}\,{{\rm{m}}^{\rm{2}}}\)
The pressure is \(P{\rm{ = 20}}\,{\rm{Pa}}\)
The force required is \(F = P \times A = 20 \times 2 = 40\;{\rm{N}}\)
Thus, the force required to get the required pressure is \({\rm{40}}\,{\rm{N}}{\rm{.}}\)

Applications of Pressure

1. The trucks carrying heavy goods have broad tyres. The broad tyres will spread the thrust exerted on the roads by the weight of the truck and hence decrease the pressure acting on the unit surface area of the road.
2. We exert more pressure while standing rather than sleeping. Our weight acting on the surface is the thrust acting on it. When we are standing, the area of the foot is the total area on which our weight is acting. Whereas while sleeping, our body is in contact with the surface which is the total area on which our weight is acting. So the pressure will decrease when we are sleeping rather than standing.
3. The foundation of a dam has a larger surface area. The bottom of the dam experiences maximum pressure due to the water-filled in it. The weight of the water in the dam exerts pressure on its bottom surface. So, to prevent damage to the bottom surface of the dam, its surface area is kept larger which helps to accommodate the weight of the water-filled in it.
4. Wide wooden sleepers are kept below the railway lines. This is because, the railway lines are narrow so when the train passes through them, there is a possibility of exerting high pressure on the ground on which they are laid on which may sink into the ground. So, to avoid any such damage, wide wooden sleepers are kept below the railway lines to increase the surface area, and these wooden sleepers help to decrease the pressure exerted on the ground.

Pressure in Liquids

Liquids above any surface exert pressure due to their weight on them. As a result, water lying in any container will exert pressure at its bottom. Thus, the pressure exerted on any surface will increase with the increase in the liquid column above it.

Look at the above bottle diagram, the lower surface will have more pressure as compared to its upper surface. This is because the pressure on any surface depends on the height of the liquid column. The height of the liquid column above the highest hole is very low, thus there is less pressure here than in the other two holes.

The pressure exerted by a liquid at any point inside the bottle can be calculated using the formula, \(P = h\rho g\)
Where
\(h\) is the height of the liquid column
\(p\) is the density of the liquid and
\(g\) is the acceleration due to gravity

Q.3. Calculate the pressure at the bottom of a container having water filled in it till the height of \({\rm{50}}\,{\rm{cm}}{\rm{.}}\) (Take \(g = 10\;{\rm{m\;}}{{\rm{s}}^{{\rm{ – 2}}}}\))
Ans: Given that,
The height of the liquid column \(h = 50\;{\rm{cm}} = 0.5\;{\rm{m}}\)
The density of water is \(\rho = 1000\;{\rm{kg\;}}{{\rm{m}}^{{\rm{ – 3}}}}\)
The acceleration due to gravity is \(g = 10\;{\rm{m\;}}{{\rm{s}}^{{\rm{ – 2}}}}\)
The pressure at the bottom of the container is
\(P = h\rho g = 0.5 \times 1000 \times 10 = 5000\,{\rm{Pa}}\)

Atmospheric Pressure

Our earth is surrounded by a thick blanket of air called the atmosphere. It starts from the surface of the earth and goes beyond \({\rm{300}}\,{\rm{km}}\) above the earth’s surface. It comprises a mixture of various gases. This layer of atmospheric gasses exerts pressure on the surface of the earth and is known as atmospheric pressure.

The atmospheric pressure can be measured by using a barometer. In a barometer, a liquid is used for measuring atmospheric pressure. Mercury is preferred as a barometric liquid and hence the atmospheric pressure is measured in terms of the height of the mercury column.

The atmospheric pressure at \(0\,^\circ {\rm{C}}\) and at sea level is \({\rm{76}}\,{\rm{cm}}\) or \({\rm{760}}\,{\rm{mm}}\) of mercury is referred to a unit called ‘atmosphere’ \(\left( {{\rm{atm}}} \right){\rm{.}}\) \({\rm{1}}\,{\rm{atm}}\) is the pressure exerted by a vertical column of mercury of \({\rm{70}}\,{\rm{cm}}\) or \({\rm{760}}\,{\rm{mm}}\) height.

\(1\;{\rm{atm}} = 1.013 \times {10^5}\;{\rm{Pa}}\)

FAQs on Pressure

The most frequently asked questions about the topic of pressure are answered here:

Q.1. What does pressure mean in science?
Ans: Pressure is the force acting normally on the unit surface area.

Q.2. What is the unit of thrust?
Ans: The unit used to measure thrust is the same as that of force. So, newton is used to measure thrust.

Q.3. Which unit is used to measure the atmospheric pressure?
Ans: The atmospheric pressure is measured using the unit atm.

Q.4. How is pressure different from force?
Ans: Pressure is the force acting perpendicularly on a unit surface area of the object whereas force is the total force acting on the object.

Q.5. How can we decrease the pressure exerted on any surface?
Ans: To decrease the pressure exerted on any surface, we need to increase the area of contact on which the force is acting.

Q.6. Does increasing the surface area increase the pressure acting on it?
Ans: No, with the increase in the surface area, the pressure exerted will decrease as the pressure is inversely proportional to the surface area on which the thrust is acting.

Learn at Embibe

Make use of the following study materials from Embibe which will definitely help you in your exams:

NCERT Solutions	NCERT Books
Class 8 Mock Test Series	Class 8 Practice Questions
Class 9 Mock Test Series	Class 9 Practice Questions
Class 10 Mock Test Series	Class 10 Practice Questions
JEE Main Mock Tests  (Class 11-12 PCM)	JEE Main Practice Questions  (Class 11-12 PCM)
NEET Mock Tests  (Class 11-12 PCB)	NEET Practice Questions  (Class 11-12 PCB)

Learn Properties of Pressure Here

Now that you are provided with all the necessary information on Pressure and we hope this detailed article is helpful to you. If you have any queries on this article or in general about Pressure, ping us through the comment box below and we will get back to you as soon as possible.